Homework$-9$: Problem $1$

$(12$ points$)$

Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test $($NOT the Limit Comparison Test.$)$ For each statement, enter C $($for "correct"$)$ if the argument is valid, or enter I $($for "incorrect"$)$ if any part of the argument is flawed. $($Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.$)$

For all and the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{1}{n^{1\mathrm{.}5}}$ converges, so by the Comparison Test, the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{ln(n)}{n^2}$ converges.

For all and the series $2{\displaystyle \underset{?}{\overset{?}{}}}\frac{1}{n^2}$ converges, so by the Comparison Test, the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{n}{n^3-9}$ converges.

For all $n>2,\frac{ln(n)}{n}>\frac{1}{n},$ and the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{1}{n}$ diverges, so by the Comparison Test, the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{ln(n)}{n}$ diverges.

For all $n>2,\frac{ln(n)}{n^2}>\frac{1}{n^2},$ and the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{1}{n^2}$ converges, so by the Comparison Test, the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{ln(n)}{n^2}$ converges.

For all and the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{1}{n^2}$ converges, so by the Comparison Test, the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{1}{n^2-6}$ converges.

For all and the series $2{\displaystyle \underset{?}{\overset{?}{}}}\frac{1}{n}$ diverges, so by the Comparison Test, the series ${\displaystyle \underset{?}{\overset{?}{}}}\frac{1}{nln(n)}$ diverges.